Target Sum

本週選做的題目也是關於動態規劃的應用，Target Sum，題目要求如下:

You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols + and -. For each integer, you should choose one from + and - as its new symbol.

Find out how many ways to assign symbols to make sum of integers equal to target S.

Example 1:

Input: nums is [1, 1, 1, 1, 1], S is 3. 
Output: 5
Explanation: 

-1+1+1+1+1 = 3
+1-1+1+1+1 = 3
+1+1-1+1+1 = 3
+1+1+1-1+1 = 3
+1+1+1+1-1 = 3

There are 5 ways to assign symbols to make the sum of nums be target 3.

我們可以以另一種方式理解這道題目，將數分為正數(P)與負數(N)，將P+N=target，所以我們有

sum(P) - sum(N) = target
sum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)
2 * sum(P) = target + sum(nums)

因此就變成尋找nums的子集P，並且有sum(P) = (sum(nums) + target) / 2，sum(nums) + target必須是偶數，在程序中多加一句半段語句sum < s || (s + sum) % 2 ? 0 : subsetSum(nums, (s + sum) / 2);

以下為我參考相關動態規劃寫法並總結出來的程序

int subsetSum(vector<int>& nums, int s) 
{
    int dp[s + 1] = { 0 };
    dp[0] = 1;
    for (int j = 0; j < nums.size(); j++)
    {
    int n = nums[j];
    for (int i = s; i >= n; i--)
    {
    dp[i] += dp[i - n];
    //cout << "dp " << i << ": " << dp[i]<<endl;
}
//cout << endl;
            
} 
    return dp[s];
}

int findTargetSumWays(vector<int>& nums, int s) 
{
    int sum = accumulate(nums.begin(), nums.end(), 0);
    return sum < s || (s + sum) % 2 ? 0 : subsetSum(nums, (s + sum) / 2); 
}